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The area of the closed region bounded by the equation $\mid x \mid+\mid y \mid= 2$ in the two-dimensional plane is 

  1. $4\pi$
  2. $4$
  3. $8$
  4. $2\pi$
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Given that, $ |x|+ |y| = 2 \quad \longrightarrow (1)$

Now, we can draw the figure.



The $ABCD $ is a rhombus with  $AC$ and $BD$ diagonals whose length are $4.$

Area of rhombus $ ABCD   = \frac{ AC\times BD}{2} = \frac{4 \times 4}{2} = 8\; \text{units}^{2}.$

Correct Answer $ : \text{C}$

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